Question

A proton is projected with a velocity of $3\times {10}^{6}m{s}^{-1}$ perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.

Answer 1

$v=3\times {10}^{6}m/s$ $B=0.6T$ $m=1.67\times {10}^{-27}kg$

$F=qvB$ $q_p=1.6\times {10}^{-19}C$

or, $\overrightarrow{a}=\frac{F}{m}=\frac{qvB}{m}$

$=\frac{1.6\times {10}^{-19}\times 3\times {10}^6\times 6\times {10}^{-1}}{1.67\times {10}^{-27}}$

$=17.245\times {10}^{13}=1.724\times {10}^{14}m/s^2$